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2 years ago

The ages of some lectures are 42,54,50,54,50,42,46,46,48 and 48.Calculate the:

Mathematics

1 answer:

gregori [183]2 years ago

7 0

Answer:

The mean age is 48

The standard deviation is 4

Step-by-step explanation:

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What is the domain and range of f(x)=(1/5)^x

maw [93]

Answer:

Domain = (-∞,∞) Range = (0,∞)

This is an exponential function.

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2 years ago

Read 2 more answers

Can someone help me pls

rodikova [14]

Answer:

C. 1/3x + 5

Step-by-step explanation:

Plugging in the values for x, we get:

y = 1/3(0) + 5 = 5

y = 1/3(6) + 5 = 7

y = 1/3(12) + 5 = 9

y = 1/3(15) + 5 = 10

I almost broke my neck trying to read this problem. :(

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2 years ago

I’m in desperate need of help

Tomtit [17]

I may not be able to help you but i do feel your pain lol

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2 years ago

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

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3 years ago

The temperature of a city at sunset was -1 overnight,the temperature decreased by 11 what was the lowest temperature overnight i

docker41 [41]

Answer:

-12

Step-by-step explanation:

-1 - 11 = -12

4 0

2 years ago